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I'm dealing with the continuity of $\sin(\frac{1}{x})$. I think that I have a proof but I'm not sure if it's right! Here is my proof:

We take the functions $g(x)=\frac{1}{x}$ and $h(x)=\sin(x)$, now we see that:

$g(x)$ is continuous in the open interval $(-\infty,0) \cup (0,\infty)$ because it's defined for every $x \in R$ except in $x=0$. On the other hand $h(x)$ is continuous all over reals. So it's also continuous in $(-\infty,0) \cup (0,\infty)$, then by the fact that the composition of two continuous functions is continuous we conclude $(hog)(x)=h(g(x))=\sin(\frac{1}{x})$ is also continuous in $(-\infty,0) \cup (0,\infty)$, so it's continuous all over reals except in $x=0$.

Am I right?! I'm not interested in other proofs because I have seen them, but I want to know if this is right! Thanks!

The very fluffy Panda
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davd
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  • You've proven that $\sin(1/x)$ is continuous at $x \neq 0$, but you still need to prove that is discontinuous at $0$. – Ben Grossmann Jun 14 '14 at 20:05
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    The function isn't defined at $x=0$ so we need not prove the discontinuity at $0$.@Omnomnomnom –  Jun 14 '14 at 20:07
  • So is this right? @user63181 – davd Jun 14 '14 at 20:17
  • @davd yes, what you've said is correct – Ben Grossmann Jun 14 '14 at 20:20
  • Yes right although your sentence "$g(x)$ is continuous in the open interval $(-\infty,0) \cup (0,\infty)$ because it's defined for every $x \in R$ except in $x=0$" isn't quite correct! @davd –  Jun 14 '14 at 20:20
  • It's not quite correct because it doesn't necessary mean that if a function is defined in that interval then immediately we conclude that it's also continuous in the same interval! Or what's the reason if this one isn't? @user63181 – davd Jun 14 '14 at 20:28
  • Yes this is what I meant @davd –  Jun 14 '14 at 20:30
  • Ok, we show that $\lim_{x_n \to x_0}{f(x_n)}$ exists and $\lim_{x_n \to x_0}{f(x_n)}=f(x_0)$ for every $x_0 \in R$ except $x_0=0$ and we end with that! Thank you! @user63181 – davd Jun 14 '14 at 20:41
  • In fact to prove the continuity of for example $x\mapsto\frac1x$ at $x_0\ne0$ we use the definition and not the sequences but it's worth to notice that this result should be shown in your course so we admit it in the exercises and it suffices to just say that this function is continuous on its domain. You're welcome @davd –  Jun 14 '14 at 20:51

2 Answers2

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The continuity of this function is clear by the composition of the continuous functions. The question is " Can we extend this function by continuity on $0$?" The answer is NO. In fact the sequence $$x_n=\frac{1}{\frac\pi2+n\pi}$$ tends to $0$ but it's image by the given function is $(-1)^n$ hasn't a limit.

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Your proof is correct; $\sin(1/x)$ is indeed continuous where it is defined.

Ben Grossmann
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